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# Mass defect phya5 Watch

1. So if we're working in u units, do we have to consider the electrons? Its mass of nucleons - mass of nucleus. When finding the mass of the nucleus do we minus the mass if the electrons from the given atom mass?

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2. (Original post by cooldudeman)
So if we're working in u units, do we have to consider the electrons? Its mass of nucleons - mass of nucleus. When finding the mass of the nucleus do we minus the mass if the electrons from the given atom mass?

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I'm pretty sure that in the exam it will be clear from the question whether or not to minus the mass of the electrons. In terms of u the mass of an electron is only 0.00054858u so won't have much effect on the total mass when a nuetron is 1.0086649u. Generally I would think that in the exam the question could say:

Find the binding energy of atom "x" using this data:
1. mass of "x" : .....u
2. mass of neutron: ......u
3. mass of proton: .....u
(4. mass of electron: ....u) (maybe)

Then the clue will be wether or not the mass of an electron is given

If the question says:
Find the mass of the nucleus of atom "x"
Then the clue to minus the mass of the electrons is the fact they used the word "nucleus" instead of "find the mass of atom "x""
3. (Original post by hello calum)
I'm pretty sure that in the exam it will be clear from the question whether or not to minus the mass of the electrons. In terms of u the mass of an electron is only 0.00054858u so won't have much effect on the total mass when a nuetron is 1.0086649u. Generally I would think that in the exam the question could say:

Find the binding energy of atom "x" using this data:
1. mass of "x" : .....u
2. mass of neutron: ......u
3. mass of proton: .....u
(4. mass of electron: ....u) (maybe)

Then the clue will be wether or not the mass of an electron is given

If the question says:
Find the mass of the nucleus of atom "x"
Then the clue to minus the mass of the electrons is the fact they used the word "nucleus" instead of "find the mass of atom "x""
Check this past question out. its where it says calculate average binding energy... in the middle. Iy gives the mass of the atom and not the nucleus. In the MS it considered the electrons. Well st least now ill be aware of this situation.
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4. (Original post by cooldudeman)
Check this past question out. its where it says calculate average binding energy... in the middle. Iy gives the mass of the atom and not the nucleus. In the MS it considered the electrons. Well st least now ill be aware of this situation.
Ah okay, so the clue to consider the electrons is the fact that they used the word "atom" in the question: "mass of Atom is ...." instead of "mass of nuceus of Atom ...."

I'm sitting this exam as well and never thought about this, thank you
5. First acknowledge that work done is FxD. Also observe that the strong nuclear force is responsible for the attractive and repulsive forces observed in nuclear fusion/fission. It is known that the strong nuclear force only acts between Hadrons (i.e. the protons/neutrons). So the work done/ energy liberated must originate from the nucleus itself. As electrons are Leptons, they cannot contribute any force to the nuclear decay and hence do not do any work.

In other words, electrons never lose mass in nuclear fusion/fission.

I can elaborate with more interesting observations should they interest you.
6. (Original post by hecandothatfromran)
First acknowledge that work done is FxD. Also observe that the strong nuclear force is responsible for the attractive and repulsive forces observed in nuclear fusion/fission. It is known that the strong nuclear force only acts between Hadrons (i.e. the protons/neutrons). So the work done/ energy liberated must originate from the nucleus itself. As electrons are Leptons, they cannot contribute any force to the nuclear decay and hence do not do any work.

In other words, electrons never lose mass in nuclear fusion/fission.

I can elaborate with more interesting observations should they interest you.
Wow that sounds like wizardry lol. No electron mass change
7. (Original post by hecandothatfromran)

In other words, electrons never lose mass in nuclear fusion/fission.
That's not really the point though. If an atom has a mass x, and we want to find its binding energy, then we still need to take into account the mass of an electron.

Mass deficit = x - n. protons*i - n. neutrons*j - n. electrons*k
Where i,j and k are the masses of a proton, neutron and electron

If we don't take of the mass of the electrons then we will get a larger binding energy.
8. (Original post by hecandothatfromran)
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Would you be able tobexplajn this diagram. I don't get the deal with thw arrows of fission and fusion. When the arrow ends, does that mean fusion/fission cant happen anymore?
Also there was this question asking why fission of heavy nucleus is likely to release more energy than a pair of light nuclei undetgoing fusion.
Is it because the fission arrow os longer?

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9. (Original post by cooldudeman)
Would you be able tobexplajn this diagram. I don't get the deal with thw arrows of fission and fusion. When the arrow ends, does that mean fusion/fission cant happen anymore?
Also there was this question asking why fission of heavy nucleus is likely to release more energy than a pair of light nuclei undetgoing fusion.
Is it because the fission arrow os longer?

Posted from TSR Mobile
It is to do with the binding energy. Fission will only release energy when the binding energy of the lighter element is greater than the binding energy of the heavier element undergoing the fission. Fusion will release energy when the binding energy of the heavier elements are greater than the binding energy of the fusing elements. This is because the binding energy is equal to the energy release when that atom is formed from all it's component nucleons. If two atoms with a low binding energy fuce to form an atom with a higher binding energy there will be a release of energy because the heavier element has a higher binding energy.

The most stable atom is the one with the highest binding energy. At this point on the diagram no energy is released by fission or fusion because the binding energy doesn't change.

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Updated: May 12, 2013
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